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I am so stuck on this question: Consider a two-asset model where asset 0 is cash, so that the price of asset 0 is $B_t=1$ for all $t \geq0$. Asset 1 has prices given by $dS_t = a(S_t) dW_t$, where the given function $a$ is positive and smooth, and such $a$ and its derivative $a'$ is bounded. Let $\xi_t$ be the time-$t$ price of a European call option with maturity $T$ and strike $K$. Let $V: [0,T] \times \mathbb{R} \rightarrow \mathbb{R}_{+}$ satisfy the PDE (with boundary condition) \begin{equation} \frac{\partial V}{\partial t} (t,S) + \frac{a(S)^2}{2} \frac{\partial^2}{\partial S^2} V(t,S) =0, \quad V(T,S)= (S-K)^{+}. \end{equation} We let $\xi_t = V(t,S_t)$ so that there is no arbitrage.

We want to show that the call option $\xi_t$ can be replicated by holding $\pi_t = U(t, S_t)$ units of stock, where $U: [0,T] \times \mathbb{R} \rightarrow \mathbb{R}$ satisfies the PDE (with boundary condition) \begin{equation} \frac{\partial U}{\partial t} (t,S) + a(S) a'(S) \frac{\partial}{\partial S} U(t,S) +\frac{a(S)^2}{2} \frac{\partial^2}{\partial S^2} U(t,S) =0, \quad U(T,S)= \mathbf{1}_{ \{S\geq K \}}. \end{equation}

What I have done so far:

Let the strategy be $\phi_t$ units of cash, $\pi_t = U(t,S_t)$ units of stock.

Clearly, by definition, $\phi_t = \xi_t - U(t,S_t) S_t$. However, this does not seem to work, as it is not self-financing:

By Ito's Lemma, $d \xi_t = d V(t, S_t) = \frac{\partial V}{\partial S} a(S_t) dW_t$ (using the first set of PDE). Hence, claiming that it is self-financing amounts to saying that \begin{equation} \frac{\partial V}{\partial S} a(S_t) dW_t = U(t,S_t) a(S_t) dW_t + \phi_t dt, \end{equation} which is clearly not true. Any ideas?

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  • $\begingroup$ Have I confused anything in this question? But the expression for $\phi_t$ just doesn't seem to work at all. $\endgroup$ – Richard Nov 30 '14 at 0:34
  • $\begingroup$ I think you are missing the quadratic variation term in Ito's lemma? Its $+\frac{1}{2}\partial_{SS}\xi_t dt$ $\endgroup$ – emcor Nov 30 '14 at 0:46
  • $\begingroup$ Ito's Lemma is not used on $V_t$ but on $\xi_t$ $\endgroup$ – emcor Nov 30 '14 at 0:49
  • $\begingroup$ I applied the Ito's lemma to $\xi_t= V(t, \xi_t)$. Is that wrong? $\endgroup$ – Richard Nov 30 '14 at 0:52
  • $\begingroup$ @emcor Also, I cancelled the quadratic variation with the first term due to the PDE. $\endgroup$ – Richard Nov 30 '14 at 0:52
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I think the title here is misleading. Let's go back to the BS world with $r=0$ to $a(S_t)=S_t \sigma.$ In that case, all you are saying is that you can replicate a call option by holding $N(d_1)$ units of stock at time $t.$

What does this have to do with the second equation? I am guessing that this is the price process of an asset of nothing option with the stock taken as numeraire so it evaluates to $N(d_1).$

So my approach to this would be to repeat the BS replication argument when $\sigma$ is allowed to be a function of $S_t.$ Then work with the stock as numeraire to get the fact that the delta satisfies the second equation.

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The exercise is really not about replicating a call with asset or nothing. It is simply about the PDE of the delta of a call.

The usual derivation of the BS equation starts by considering a portfolio short the option $$ \Pi_t = \delta^0_t B_t + \delta_t S_t - V(t,S_t) $$ Assuming the portfolio is selfinancing (and interest rate = 0), we get $$ d\Pi_t = \delta_t dS_t - dV(t,S_t) = (\delta_t - \partial_SV)a(S_t)dW_t - (\partial_tV + \frac{1}{2}a(S_t)^2\partial^2_{SS}V)dt $$ We make the portfolio riskless by choosing $\delta_t = \partial_S V(t,S_t)$. And as $\Pi$ is now riskless it must earn the risk free rate of 0 so that gives us the PDE for V. $$ \partial_tV + \frac{1}{2}a(S)^2\partial^2_{SS}V = 0 $$ Deriving the PDE for V wrt S gives the PDE for $\delta(t,S) = \partial_S V(t,S) = U(t,S)$: $$ \partial_tU + a(S)a'(S)\partial_{S}U + \frac{1}{2}a(S)^2\partial^2_{SS}U = 0 $$ Setting the initial value $\Pi_0 = 0$ ensures that $\Pi_t = 0$ $\forall t$ so we replicate $V$ perfectly by holding $\delta_t = U(t,S_t)$ units of $S_t$.

PS: I wrote $\delta^0_t$ and $\delta_t$ instead of $\phi_t,\pi_t$. The mistake in your self-financing equation is that it should write $$ dV(t,S_t) = \pi_tdS_t + \phi_t dB_t $$ but $B_t = 1$ so $dB_t = 0$ and we are left with $$ \partial_S V(t,S_t) a(S_t)dW_t = U(t,S_t)a(S_t)dW_t $$ which is another way of finding $\pi_t = U(t,S_t) = \partial_S V(t,S_t)$.

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Without getting into all the Math one thing should be clear that: Call option is equivalent to: long asset or nothing AND short cash or nothing options.

You cannot replicate a call option without asset or nothing since replicating portfolio for long call requires holding N(d1) quantity of the underlying. Asset or nothing gives you this exposure directly. Shorting cash or nothing which pays $K at maturity provides the remaining exposure required for replicating the long call option.

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